Homework # 2

5.1 By monotonicity,

P{|X | ≥ a} ≤ P{g(|X |) ≥ g(a)} ≤Eg(|X |)

g(a)

5.16. The correct form should be

P{X > i + j|X ≥ i} = P{X > j}

Indeed,

P{X > j} =∞∑

k=j+1

(1 − p)kp = (1 − p)j+1p

∞∑

k=0

(1 − p)k = (1 − p)j+1

Replacing j by i + j,P{X > i + j} = (1 − p)i+j+1

Replacing j by i − 1,P{X ≥ i} = P{X > i − 1} = (1 − p)i

Thus

P{X > i + j|X ≥ i} =P{X > i + j, X ≥ i}

P{X ≥ i}

=P{X > i + j}

P{X ≥ i}= (1 − p)j+1 = P{X > j}

5.19. Intuitively, this follows from linearty of the expectation. On the other hand,justification is needed for the inifinite summation. It is not hard if the monotonic convern-gence theorem (will be taught later) is allowed. Without monotonic converngence theorem,we need a delicate algebrac operation.

Notice that the random variable

X =

∞∑

n=1

1An

represents the number of hapenning events among A1, · · · , An, · · ·. By definition,

E

( ∞∑

n=1

1An

)

= E(X) =

∞∑

k=0

kP{X = k} =

∞∑

k=1

kP{X = k}

For any finite set I ⊂ {1, 2, · · ·}, set

CI =(

⋂

n∈I

An

)

∩(

⋂

n6∈I

Acn

)

1

Then for any finite sets I, J ⊂ {1, 2, · · ·} with I 6= J , CI ∩ CJ = φ.

In addition, for any n ≥ 1,

⋃

finite I: n∈I

CI = An

Further, for any k ≥ 1, by countable additivity of the probability

P{X = k} = P

(

⋃

I: #(I)=k

CI

)

=∑

I: #(I)=k

P (CI)

Hence,

E

( ∞∑

n=1

1An

)

=∞∑

k=1

k∑

I: #(I)=k

P (CI)

Notice thatk

∑

I: #(I)=k

P (CI) =∑

I: #(I)=k

∑

n: n∈I

P (CI)

Hence,

E

( ∞∑

n=1

1An

)

=∞∑

k=1

∑

I: #(I)=k

∑

n: n∈I

P (CI) =∞∑

n=1

∑

finite I: n∈I

P (CI)

=

∞∑

n=1

P

(

⋃

finite I: n∈I

CI

)

=

∞∑

n=1

P (An)

where the third equality follows from countable additivity of probability.

5.20. Notice that

X =

∞∑

n=0

1{X>n}

The requested conclusion follows from the conclusion of 5.19.

Alternative proof

E(X) =∞∑

k=1

kP{X = k} =∞∑

k=1

k(

P{X > k − 1} − P{X > k})

By the fact

n∑

k=1

k(

P{X > k − 1} − P{X > k})

=

n−1∑

k=0

(k + 1)P{X > k} −

n∑

k=1

kP{X > k}

= −nP{X > n} +n−1∑

k=0

P{X > k}

2

Whn n → ∞, the both sides converges or diverges simultaneously. When converging, theleft hand gives that E(X) < ∞. By (the spirit of )Markov inequality,

nP{X > n} ≤∑

k=n+1

kP{X = k} −→ 0 (n → ∞)

So we have

E(X) =∞∑

k=0

P{X > k}

3